| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | 3x3 Matrices |
| Type | Find inverse then solve system |
| Difficulty | Standard +0.3 Part (a) requires computing a 3×3 determinant to check for unique solutions—a standard procedure. Part (b) involves solving a 3×3 system using Gaussian elimination or matrix methods, which is routine for Further Maths students. Both parts are textbook exercises requiring careful calculation but no novel insight or complex problem-solving. |
| Spec | 4.03l Singular/non-singular matrices4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| METHOD 1: \(\det = 2(12 + 16) + 7(0 - 14) + 2(0 + 21) = 0\) | M1A1 E1 | FT det |
| \(\therefore\) the equations do not have a unique solution | ||
| METHOD 2: Reduction to echelon form | (M1) | |
| Correct matrix to determine nature of solutions, eg \(\begin{pmatrix}2 & -7 & 2 & | & a \\ 0 & 33 & -22 & | & 11b \\ 0 & 33 & -22 & |
| Correct statement: eg No unique solutions when \(11b = -7a - 2c\) | (E1) |
| Answer | Marks | Guidance |
|---|---|---|
| METHOD 1: \(\begin{pmatrix}1 & 8 & -6 & | & 5 \\ 2 & 4 & 6 & | & -3 \\ -5 & -4 & 9 & |
| By row operations: \(\begin{pmatrix}1 & 8 & -6 & | & 5 \\ 0 & -12 & 18 & | & -13 \\ 0 & 36 & -21 & |
| Then \(\begin{pmatrix}1 & 8 & -6 & | & 5 \\ 0 & -12 & 18 & | & -13 \\ 0 & 0 & 33 & |
| Therefore: \(z = \frac{-7}{11}\), \(y = \frac{17}{132}\), \(x = \frac{5}{33}\) cao | A2 | A1 for any 2 correct |
| METHOD 2: Rearranging \(x = 5 - 8y + 6z\) and Substituting: 2nd equation: \(-12y + 18z = -13\); 3rd equation: \(36y - 21z = 18\) | (M1) | |
| Solving, \(z = \frac{-7}{11}\), \(y = \frac{17}{132}\), \(x = \frac{5}{33}\) cao | (A1) (m1) (A2) | A1 for any 2 correct |
| METHOD 3: Let \(A = \begin{pmatrix}1 & 8 & -6 \\ 2 & 4 & 6 \\ -5 & -4 & 9\end{pmatrix}\) and \(B = \begin{pmatrix}5 \\ -3 \\ -7\end{pmatrix}\) | ||
| \(\det = 1(36 + 24) - 8(18 + 30) - 6(-8 + 20) = -396\) | (M1) M0 no working | (A1) |
| \(A^{-1} = \frac{1}{-396}\begin{pmatrix}60 & -48 & 72 \\ -48 & -21 & -18 \\ 12 & -36 & -12\end{pmatrix}\) | ||
| Therefore, \(X = A^{-1}B\) | ||
| \(X = \frac{1}{-396}\begin{pmatrix}60 & -48 & 72 \\ -48 & -21 & -18 \\ 12 & -36 & -12\end{pmatrix}\begin{pmatrix}5 \\ -3 \\ -7\end{pmatrix}\) | (m1) | |
| \(X = \begin{pmatrix}5/33 \\ 17/132 \\ -7/11\end{pmatrix}\) cao | (A2) | A1 for any 2 correct |
## Part a)
METHOD 1: $\det = 2(12 + 16) + 7(0 - 14) + 2(0 + 21) = 0$ | M1A1 E1 | FT det
$\therefore$ the equations do not have a unique solution | | |
METHOD 2: Reduction to echelon form | (M1) | |
Correct matrix to determine nature of solutions, eg $\begin{pmatrix}2 & -7 & 2 & | & a \\ 0 & 33 & -22 & | & 11b \\ 0 & 33 & -22 & | & -7a - 2c\end{pmatrix}$ | (A1) | |
Correct statement: eg No unique solutions when $11b = -7a - 2c$ | (E1) | |
## Part b)
METHOD 1: $\begin{pmatrix}1 & 8 & -6 & | & 5 \\ 2 & 4 & 6 & | & -3 \\ -5 & -4 & 9 & | & -7\end{pmatrix}$ | M1 |
By row operations: $\begin{pmatrix}1 & 8 & -6 & | & 5 \\ 0 & -12 & 18 & | & -13 \\ 0 & 36 & -21 & | & 18\end{pmatrix}$ | A1 | Reduction to $\begin{pmatrix}1 & & \\ 0 & & \\ 0 & & \end{pmatrix}$
Then $\begin{pmatrix}1 & 8 & -6 & | & 5 \\ 0 & -12 & 18 & | & -13 \\ 0 & 0 & 33 & | & -21\end{pmatrix}$ | A1 | Reduction to $\begin{pmatrix}1 & & \\ 0 & & \\ 0 & 0 & \end{pmatrix}$
Therefore: $z = \frac{-7}{11}$, $y = \frac{17}{132}$, $x = \frac{5}{33}$ cao | A2 | A1 for any 2 correct
METHOD 2: Rearranging $x = 5 - 8y + 6z$ and Substituting: 2nd equation: $-12y + 18z = -13$; 3rd equation: $36y - 21z = 18$ | (M1) | |
Solving, $z = \frac{-7}{11}$, $y = \frac{17}{132}$, $x = \frac{5}{33}$ cao | (A1) (m1) (A2) | A1 for any 2 correct
METHOD 3: Let $A = \begin{pmatrix}1 & 8 & -6 \\ 2 & 4 & 6 \\ -5 & -4 & 9\end{pmatrix}$ and $B = \begin{pmatrix}5 \\ -3 \\ -7\end{pmatrix}$ | |
$\det = 1(36 + 24) - 8(18 + 30) - 6(-8 + 20) = -396$ | (M1) M0 no working | (A1) |
$A^{-1} = \frac{1}{-396}\begin{pmatrix}60 & -48 & 72 \\ -48 & -21 & -18 \\ 12 & -36 & -12\end{pmatrix}$ | | |
Therefore, $X = A^{-1}B$ | | |
$X = \frac{1}{-396}\begin{pmatrix}60 & -48 & 72 \\ -48 & -21 & -18 \\ 12 & -36 & -12\end{pmatrix}\begin{pmatrix}5 \\ -3 \\ -7\end{pmatrix}$ | (m1) |
$X = \begin{pmatrix}5/33 \\ 17/132 \\ -7/11\end{pmatrix}$ cao | (A2) | A1 for any 2 correct\begin{enumerate}[label=(\alph*)]
\item Determine whether or not the following set of equations
$$\begin{pmatrix} 2 & -7 & 2 \\ 0 & 3 & -2 \\ -7 & 8 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$$
has a unique solution, where $a$, $b$, $c$ are constants. [3]
\item Solve the set of equations
\begin{align}
x + 8y - 6z &= 5, \\
2x + 4y + 6z &= -3, \\
-5x - 4y + 9z &= -7.
\end{align}
Show all your working. [5]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2019 Q3 [8]}}